dongyan7950 2018-07-09 04:26
浏览 53

未定义的变量:C:\ xampp \ htdocs \ graceillam.com \ database.php中的POST_

Here i'm using html code to insertion a data to phpmyadmin database,i'm getting error on database.php file.

Here i attaching code details and error details also.

html file:-

 <form action="database.php" method="POST">
              <fieldset>
                <input type="text" name="name" id="name" value="" class="input-block-level" placeholder="Name" />
                <input type="text" name="email" id="email" value="" class="input-block-level" placeholder="Email" />
                <input type="date" name="adt" id="adt" value="" class="input-block-level" placeholder="Arrival Date" />
                  <input type="text" name="num" id="num" value="" class="input-block-level" placeholder="PhoneNumber" />
                  <textarea rows="11" name="message" id="message" class="input-block-level" placeholder="Comments"></textarea>
                 <div class="row-fluid">  
                  <div class="span2" align=top> 
                  <div class="g-recaptcha" data-sitekey="6Ld4bgcTAAAAAJL6HujpsmyQvn5sJzHobWY-XdNo"></div>  

                    </div>  
                   <div class="span10 actions" align=top> 
                     <input align=left type="submit" value="Send Your Message" name="submit" id="submitButton" class="btn btn-info pull-right" title="Click here to submit your message!" />
                      <span id="captcha" style="margin-left:100px;color:red" /> 
                   </div> 
                </div>
          </div>
          </fieldset>
          </form>

2) php file:-

<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "phpmyadmin";

// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
    die("Connection failed: " . $conn->connect_error);
} 

$sql = "INSERT INTO user (name,email,adt,num,message)
VALUES ('$POST_[name],'$POST_[email]','$POST_[adt]','$POST_[num]','$POST_[message]')";

if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
} else {
    echo "Error: " . $sql . "<br>" . $conn->error;
}

$conn->close();
?>

3)error:-

Notice: Undefined variable: POST_ in C:\xampp\htdocs\graceillam.com\database.php on line 15

Notice: Undefined variable: POST_ in C:\xampp\htdocs\graceillam.com\database.php on line 15

Notice: Undefined variable: POST_ in C:\xampp\htdocs\graceillam.com\database.php on line 15

Notice: Undefined variable: POST_ in C:\xampp\htdocs\graceillam.com\database.php on line 15

Notice: Undefined variable: POST_ in C:\xampp\htdocs\graceillam.com\database.php on line 15
Error: INSERT INTO user (name,email,adt,num,message) VALUES (','','','','')
You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near '','','','','')' at line 2

This error showing when we insert the data from the html file and the data will be connecting to database via.

  • 写回答

1条回答 默认 最新

  • dongpan1416 2018-07-09 04:32
    关注

    Check you syntax. $POST_ was supposed to be $_POST and add ' in every index since $_POST was an associative array.

    <?php
    $servername = "localhost";
    $username = "root";
    $password = "";
    $dbname = "phpmyadmin";
    
    // Create connection
    $conn = new mysqli($servername, $username, $password, $dbname);
    // Check connection
    if ($conn->connect_error) {
        die("Connection failed: " . $conn->connect_error);
    } 
    
    $sql = "INSERT INTO user (name,email,adt,num,message)
    VALUES ('$_POST['name'],'$_POST['email']','$_POST['adt']','$_POST['num']','$_POST['message']')";
    
    if ($conn->query($sql) === TRUE) {
    echo "New record created successfully";
    } else {
        echo "Error: " . $sql . "<br>" . $conn->error;
    }
    
    $conn->close();
    ?>
    
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